Well (^) is already used for their traditional meaning and using this exact operator would require 1. Shadowing (^) from prelude 2. Making (a -> a) an instance of Num (impossible to do sanely) You can just use a different operator f .^. n = foldl (.) id $ replicate n f main = print . (+1) .^. 5 $ 1 Will print 6 Cheers, Danny Gratzer On Tue, Dec 10, 2013 at 10:45 AM, Doug McIlroy wrote:

Is there a trick whereby the customary notation f^n for iterated functional composition ((\n f -> foldl (.) id (replicate n f)) n f) can be defined in Haskell?

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